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For fixed $n \in \mathbb{N}$ consider integer solutions to $$x^3+y^3+z^3=n \qquad (1) $$

If $n$ is a cube or twice a cube, identities exist.

Elkies suggests no other polynomial identities are known.

For which $n$ (1) has infinitely many integer solutions?

Added

Is there $n$, not a cube or twice a cube, which allows infinitely many solutions?

Added 2019-09-23:

The number of solutions can be unbounded.

For integers $n_0,A,B$ set $z=Ax+By$ and consider $x^3+y^3+(Ax+By)^3=n_0$. This is elliptic curve and it may have infinitely many rational points coming from the group law. Take $k$ rational points $(X_i/Z_i,Y_i/Z_i)$. Set $Z=\rm{lcm}\{Z_i\}$.

Then $n_0 Z^3$ has the $k$ integer solutions $(Z X_i/Z_i,Z Y_i/Z_i)$.

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    $\begingroup$ Let $n = 1, 2$ and we have solutions for $x, y, z$ with a value $t \in \mathbb{N}$ however not all $x, y, z$ are positive, though I think this may serve as a little bit of help: $$\begin{align} 1 &= (9t^3 + 1)^3 + (9t^4)^3 + (-9t^4 - 3t)^3 \\ 2 &= (6t^3 + 1)^3 + (-6t^3 - 1)^3 + (-6t^2)^3 \end{align}$$ or for big solutions for $n = 1$: $$1 = (1 - 9t^3 + 648t^6 + 3888t^9)^3 + (-135t^4 + 3888t^{10})^3 + (3t - 81t^4 - 1296t^7 - 3888t^{10})^3$$ $\endgroup$
    – Mr Pie
    Commented Oct 27, 2017 at 5:38
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    $\begingroup$ The expression for n = 2 should read: $(6t^3 + 1)^3 + (- 6t^3 + 1)^3 + (- 6t^2)^3 = 2$ That is, the 2nd sign in the 2nd pair of parentheses should be + rather than -. $\endgroup$
    – Chris Manning
    Commented Nov 26, 2017 at 5:37
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    $\begingroup$ @joro: Using Huisman's results, one finds that $$x^3+y^3+z^3 = 972$$ already has $96$ solutions in the "small" finite range that he searched. Similarly for other $n$. It is tempting to speculate that, if the range were infinite, then it and others in fact have infinitely many solutions like $n=1$. $\endgroup$
    – Tito Piezas III
    Commented Dec 30, 2017 at 3:17
  • $\begingroup$ @TitoPiezasIII What if in general the solutions are exponentially growing? $\endgroup$
    – joro
    Commented Dec 30, 2017 at 9:15
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    $\begingroup$ The Wikipedia article Sums of three cubes contains some results on this and also quite long list of references. $\endgroup$
    – Martin Sleziak
    Commented Sep 18, 2019 at 5:42

3 Answers 3

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For $n\equiv \pm 4\pmod{9}$ there is no solution to $(1)$. Otherwise, for $n\ge 1$, it is conjectured that there are always solutions, even infinitely many. There are no analytic results, but heuristics suggest that given $n$, not $0$ or $\pm 4\pmod{9}$, solutions should occur infinitely often, asymptotically $c\log(N)$ solutions in $|x|,|y|,|z|<N$, see papers of Conn and Vaserstein.

The topic has been discussed quite frequently, see also sum of three cubes and parametric solutions, Are nontrivial integer solutions known for $x^3+y^3+z^3=3$?, Efficient computation of integer representation as a sum of three squares, etc. For a collection on polynomial parametric solutions, see https://sites.google.com/site/tpiezas/010.

See also the Euler-Binet solutions to $x^3+y^3=z^3+w^3$, $$ x = 1 − (p − 3q)(p^2 + 3q^2), $$ $$ y = −1 + (p + 3q)(p^2 + 3q^2), $$ $$ z = (p + 3q) − (p^2 + 3q^2)^2, $$ $$ w = −(p − 3q) + (p^2 + 3q^2)^2. $$

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  • $\begingroup$ Thank you. So unless n is a cube or twice a cube no infinite solutions are known? $\endgroup$
    – joro
    Commented Aug 8, 2013 at 8:58
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As stated by Dietrich Burde, it is known that there is no solution for $n\equiv \pm 4 \pmod{9}$, and conjectured that there are infinitely many solutions otherwise.

A cryptic aspect is that it is not even known that there exists one solution for all $n \not\equiv \pm 4 \pmod{9}$.
Today the smallest number for which the problem is open is $n=114$.

Here is a (non-exhaustive) history of the latest solutions found for $n \le 100$ (see here and there):

(1960s)

  • $87 = 4271^3 – 4126^3 – 1972^3$
  • $96 = 13139^3 -15250^3 + 10853^3$
  • $91 = 83538^3 – 67134^3 – 65453^3$
  • $80 = 103532^3 -112969^3 + 69241^3$

(1990s)

  • $39 = 134476^3 - 159380^3 + 117367^3$
  • $75 = 435203083^3 – 435203231^3 + 4381159^3$
  • $84 = 41639611^3 – 41531726^3 – 8241191^3$

(2000s)

  • $30 = 2220422932^3 – 2218888517^3 – 283059965^3$
  • $52 = 23961292454^3 - 61922712865^3 + 60702901317^3$
  • $74 = 66229832190556^3 − 284650292555885^3 + 283450105697727^3$

(2019)

  • $33 = 8866128975287528^3 - 8778405442862239^3 -2736111468807040^3$
  • $42 = 80435758145817515^3 - 80538738812075974^3 + 12602123297335631^3$
  • $3 = 569936821221962380720^3 - 569936821113563493509^3 - 472715493453327032^3$
  • $906 = 72054089679353378^3 -74924259395610397^3 + 35961979615356503^3$
  • $165 = 383344975542639445^3 -385495523231271884^3 + 98422560467622814^3$

Remark: for $n \le 1000$, the problem is still open only for $114$, $390$, $579$, $627$, $633$, $732$, $921$, and $975$ (see this paper and this paper, and also this).

Numberphile's videos:

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    $\begingroup$ The number 33 has now been expressed as a sum of 3 cubes. See pub.ist.ac.at/~tbrownin. $\endgroup$
    – KConrad
    Commented Mar 8, 2019 at 21:14
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    $\begingroup$ $$33=(8866128975287528)^3+(-8778405442862239)^3+(-2736111468807040)^3$$ According to this address, the discoverer should be Tim Browning. $\endgroup$
    – Sebastien Palcoux
    Commented Mar 8, 2019 at 22:17
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    $\begingroup$ It was actually Andrew Booker in Bristol that found this solution! $\endgroup$
    – Gil Kalai
    Commented Mar 10, 2019 at 18:52
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    $\begingroup$ The number 42 is now known to be a sum of three cubes: $(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3$. I found this on Drew Sutherland's website. $\endgroup$
    – KConrad
    Commented Sep 6, 2019 at 0:10
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    $\begingroup$ I heard back from Drew. He and Booker worked together on this, and the change of their websites to show only the representation of 42 was inspired by how Browning made Booker's result for $33$ known. $\endgroup$
    – KConrad
    Commented Sep 6, 2019 at 2:02
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https://sites.google.com/site/tpiezas/010 cites the following polynomial identity $$(m^3-3^6n^9)^3+(-m^3+3^5mn^6+3^6n^9)^3+ (3^3m^2n^3+3^5mn^6)^3=m(3^2m^2n^2+3^4mn^5+3^6n^8)^3$$ valid for any $n$ and $m$.

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  • $\begingroup$ Thanks. But the RHS varies as m,n vary so this doesn't represent fixed n? $\endgroup$
    – joro
    Commented Aug 8, 2013 at 9:56
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    $\begingroup$ Yes, you are right. This identity does not produce infinitely many integer solutions. It only proves Ryley's Theorem that "any non-zero rational number $m$ is the sum of three rational cubes in an infinite number of non-trivial ways.” $\endgroup$
    – Zurab Silagadze
    Commented Aug 8, 2013 at 10:08

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